On his way home from the laboratory, Louie realized that he left a test tube containing $2560$ yeast cells in the lab. Every two minutes, the number of cells in the test tube increases by $50\%$. If the number of cells reaches $98{,}415$, the test tube will explode! Naturally, he turned around and rushed back to the lab. It took Louie $t$ minutes to return to the lab, and he found the test tube intact. Write an inequality in terms of $t$ that models the situation.
The strategy This problem involves the number of yeast cells in a test tube increasing by $50\%$ every $2$ minutes. So, to find the number of yeast cells in the test tube over time, we repeatedly multiply the original number of cells, $2560$, by $1.5$. [Why?] Because of this, we know we can model the situation with an exponential expression of the form $ab^x$, where $a$ is $2560$ and $b$ is $1.5$. We now only need to find $x$, which represents the number of times the cells have increased by $50\%$. Finding the exponent Suppose $t$ minutes have passed. Since the number of yeast cells increase by $50\%$ every two minutes, in $t$ minutes, the number of yeast cells will increase by $50\%$ a total of $\dfrac{t}{2}$ times. Writing an inequality We can now replace $x$ in the original model with $\dfrac{t}{2}$. Therefore, the expression 2560 ⋅ ( 1.5 ) t 2 2560\cdot (1.5)\^{\frac{t}{2}} models the number of yeast cells in the test tube after $t$ minutes, or the number of yeast cells in the test tube when Louie returned to the lab. Since the test tube was still intact when Louie returned to the lab, we know that the number of yeast cells in the test tube must be less than $98{,}415$. Therefore, we have the following inequality. 2560 ⋅ ( 1.5 ) t 2 < 98,415 2560\cdot (1.5)\^{\frac{t}{2}}<98{,}415 The answer An inequality that models the situation is 2560 ⋅ ( 1.5 ) t 2 < 98415 2560\cdot (1.5)\^{\frac{t}{2}}<98415.